Q. A photon of energy $E$ ejects a photoelectrons from a metal surface whose work function is $W_{0}.$ If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r,$ then the radius $r,$ is given by, (in the usual notation)
A) $\sqrt{\frac{2m(E - W_{0})}{eB}}$
B) $\sqrt{2m(E - W_{0})eB}$
C) $\frac{\sqrt{2e(E - W_{0})}}{mB}$
D) $\frac{\sqrt{2m(E - W_{0})}}{eB}$
Explanation
(d) From Einstein equation $E = W_{0} + \frac{1}{2}mv^{2}$
$\sqrt{\frac{2(E - W_{0})}{m}} = v$
and a charged particle placed in uniform magnetic field experience a force
$F = \frac{mv^{2}}{r} \Rightarrow evB = \frac{mv^{2}}{r} \Rightarrow r = \frac{mv}{eB}$
$\Rightarrow r = \frac{\sqrt{2m(E - W_{0})}}{eB}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.